A cosmic ray electron moves at\({\rm{7}}{\rm{.50 \times 1}}{{\rm{0}}^{\rm{6}}}{\rm{ m/s}}\)perpendicular to the Earth’s magnetic field at an altitude where field strength is\({\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ T}}\). What is the radius of the circular path the electron follows?

When a charged particle moves in the presence of a magnetic field, a force acts on the mong charge particle given by Fleming’s right-hand rule and which deviates the path of the moving charged particle.

To obtain the circular path that the electron follows.

The equation used is:\({\rm{r = }}\frac{{{\rm{mv}}}}{{{\rm{qB}}}}\).

Solving it for the value of\({\rm{r}}\)where the value of\({\rm{m}}\)is the mass of the electron.

\(\begin{array}{c}{\rm{r}} = \frac{{{\rm{mv}}}}{{{\rm{qB}}}}\\ = \frac{{\left( {{\rm{9}}{\rm{.1 \times 1}}{{\rm{0}}^{{\rm{ - 31}}}}\;{\rm{kg}}} \right){\rm{ \times }}\left( {{\rm{7}}{\rm{.50 \times 1}}{{\rm{0}}^{\rm{6}}}{\rm{ m/s}}} \right)}}{{\left( {{\rm{1}}{\rm{.6 \times 10}}{}^{{\rm{ - 19}}}\;{\rm{C}}} \right){\rm{ \times }}\left( {{\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ T}}} \right)}}\\ = {\rm{4}}{\rm{.27 m}}\end{array}\)

Therefore, the circular path the electron follows has obtained a path of measurement \({\rm{4}}{\rm{.27 m}}\).